Problem 33
Question
Graph the function. $$y=\ln \left(x^{2}+1\right)$$
Step-by-Step Solution
Verified Answer
The function \(y=\ln \left(x^{2}+1\right)\) is graphed by first determining the function’s domain (all real numbers), identifying the y-intercept (0), recognizing that the function is even and symmetric around the y-axis, and finally sketching the graph while noting it has a minimum point at the origin and a horizontal asymptote at y=-infinity.
1Step 1 : Establish the Domain of the Function
In this step, determine the domain of the function, which includes all x values for which the function is defined. Since this function is \(\ln \left(x^{2}+1\right)\), the domain is all real numbers. This is because the square of any real number is non-negative, thus the expression \(x^{2}+1\) is always positive, allowing the natural logarithm to be taken.
2Step 2: Compute the y-intercept
The y-intercept is the point where the function crosses the y-axis and is found by setting \(x=0\). Here, substitute x in \(\ln \left(x^{2}+1\right)\) with 0. \(y = \ln \left((0)^{2}+1\right) = \ln(1) = 0\). So, the y-intercept is \(y=0\).
3Step 3: Identify Symmetry of the Function
The given function is even because replacing \(x\) with \(-x\) leaves the expression \(x^{2}+1\) unchanged. Thus, the graph of the function is symmetric with respect to the y-axis. Knowing this will reduce the workload as only half the graph (either to the left or right of y-axis) needs to be sketched, and the other half copied symmetrically.
4Step 4: Sketch the Graph
Considering the facts that \(\ln(1)=0\) which gives minimum point at x = 0 and the function has a horizontal asymptote at y=-infinity, start sketching from the origin (minimum point). Notice from the square operation that as x diverges from zero in either positive or negative direction, y values will increase, and there will also be symmetry around the y-axis. This completes the sketch of the graph.
Key Concepts
Function DomainY-InterceptSymmetryGraph Sketching
Function Domain
The domain of a function is the set of all possible input values \(x\) for which the function is defined. For the function \(y = \ln(x^2 + 1)\), the domain includes all real numbers. This is because \(x^2 + 1\) is always greater than zero, allowing us to safely take the natural logarithm.
Often when dealing with logarithmic functions, the challenge is ensuring the expression is positive. But here, since \(x^2\) is always non-negative and \(1\) ensures it's never zero, the domain is comfortably all real numbers.
Often when dealing with logarithmic functions, the challenge is ensuring the expression is positive. But here, since \(x^2\) is always non-negative and \(1\) ensures it's never zero, the domain is comfortably all real numbers.
Y-Intercept
The y-intercept of a graph is where the function crosses the y-axis, occurring when \(x = 0\). For the function \(y = \ln(x^2 + 1)\), substituting \(x = 0\) gives us \(y = \ln(1) = 0\).
This means the graph crosses the y-axis at the origin \((0,0)\). Finding the y-intercept helps anchor the graph, giving you a clear starting point when sketching or analyzing the function.
This means the graph crosses the y-axis at the origin \((0,0)\). Finding the y-intercept helps anchor the graph, giving you a clear starting point when sketching or analyzing the function.
Symmetry
Symmetry in functions can simplify graphing by reducing the amount of work needed. If a function is symmetrical, you only need to graph one part and then reflect it.
For \(y = \ln(x^2 + 1)\), replacing \(x\) with \(-x\) yields the same function because \((-x)^2 = x^2\). This shows the function is even, indicating symmetry about the y-axis. Understanding symmetry can enhance your approach to solving and graphing functions efficiently.
For \(y = \ln(x^2 + 1)\), replacing \(x\) with \(-x\) yields the same function because \((-x)^2 = x^2\). This shows the function is even, indicating symmetry about the y-axis. Understanding symmetry can enhance your approach to solving and graphing functions efficiently.
Graph Sketching
Graph sketching involves translating the function's properties onto a visual representation. For \(y = \ln(x^2 + 1)\), start by plotting the y-intercept, \((0, 0)\). From this point, observe that as \(x\) moves away from zero, either positively or negatively, the value of the function increases.
There are no real limits on \(x\), thus the graph extends infinitely in either direction. This function's even symmetry allows mirroring about the y-axis, simplifying the sketching process.
There are no real limits on \(x\), thus the graph extends infinitely in either direction. This function's even symmetry allows mirroring about the y-axis, simplifying the sketching process.
- The graph has a minimum at the origin.
- It never actually touches the x-axis but stretches upwards.
Other exercises in this chapter
Problem 32
Use the Integral Mean Value Theorem to estimate the value of the integral. $$\int_{-1}^{1} \frac{3}{x^{3}+2} d x$$
View solution Problem 33
Evaluate the definite integral. $$\int_{-1}^{1} \frac{x}{\left(x^{2}+1\right)^{2}} d x$$
View solution Problem 33
The velocity of an object at various times is given. Use the data to estimate the distance traveled. $$\begin{array}{|l|r|r|r|r|r|r|r|} \hline t(\mathrm{s}) & 0
View solution Problem 33
Find an equation of the tangent line at the given value of \(x\) $$y=\int_{0}^{x} \sin \sqrt{t^{2}+\pi^{2}} d t, x=0$$
View solution